Convergence of a spiral in $\mathbb{C}$

Question

Does the series $$\sum_{k=0}^{\infty}\frac{i^k}{k!}$$converge, and if so, what is the value of it?

Do you know the Taylor expansion of $\exp (x)$? – Fabian Jan 02 '13 at 19:14 — Fabian, Jan 02 '13 at 19:14
Yes, for some reason I forgot it. Thank you all. – Meow Jan 02 '13 at 19:15 — Meow, Jan 02 '13 at 19:15
This question is so trivial I think I may delete it. – Meow Jan 02 '13 at 19:16 — Meow, Jan 02 '13 at 19:16

score 5 · Answer 1 · answered Jan 02 '13 at 19:14

5

Hint: $$e^z=\sum_{k=0}^{\infty}\frac{z^k}{k!}$$

answered Jan 02 '13 at 19:14

Nameless

13,456

2

That seems more like an answer than a hint. – robjohn Jan 02 '13 at 19:56
2

@robjohn Sometimes the line between answers and hints can be very subtle. This is one of them. – Nameless Jan 02 '13 at 19:59

score 4 · Answer 2 · edited Jan 02 '13 at 19:50

Well, first of all: what's that power series' convergence radius? $$a_k:=\frac{i^k}{k!}\Longrightarrow \frac{a_{k+1}}{a_k}=\frac{i^{k+1}}{(k+1)!}\frac{k!}{i^k}=\frac{i}{k+1}\xrightarrow[k\to\infty]{}0$$

Thus, the series converges for all $\,i\in\Bbb R\,$

Advice: Don't use the letter $\,i\,$ as it usually stands for $\,i=\sqrt{-1}\,$ in mathematics..unless you really meant this $\,i\,$ , of course.

score 3 · Answer 3 · answered Jan 02 '13 at 19:14

3

The sum of the series is $\exp{(i)}$, which has the value of $\cos{1} + i \sin{1}$ (the arguments being in radians).

answered Jan 02 '13 at 19:14

Ron Gordon

138,521

amWhy · Answer 4 · 2013-01-03T17:50:30.350

3

Recall: $\;\;$for $\large\;x \in \mathbb{C}\,:$ $\large\;\;\;\displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!} = e^x$

In your case, we have $\large x = i,\;$ giving us $\large \;e^x = e^i = e^{i\theta},\text{ where}\;\;\theta = 1$.
Now recall Euler's Formula: $$\large \;\;e^{i\theta} = \cos \theta + i \sin\theta,$$
$\quad$ and simply evaluate at $\;\large\theta = 1$.

edited Jan 03 '13 at 17:50

answered Jan 02 '13 at 19:18

amWhy

209,954

and $e^i=\cos(1)+i\sin(1)$. – pshmath0 Jan 02 '13 at 20:09
Alyosha: is this clear now? Just checking. $\quad$:-) – amWhy Jan 03 '13 at 18:04
yes, much better. – pshmath0 Jan 07 '13 at 10:25

score 1 · Answer 5 · answered Jan 02 '13 at 20:59

Alternately (and equivalently to several of the other answers), if you didn't know the Euler formula but did know the power series for sin and cos, you could reason as follows:

$i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=i^0=1$. Therefore, the numerators of the power series are periodic of period 4; what's more, they split off naturally into a real part (the even members) and an imaginary part (the odd members): $$\begin{align*} \sum_{k=0}^{\infty}\frac{i^k}{k!} &= 1+\frac{i}{1!}+\frac{i^2}{2!}+\frac{i^3}{3!}+\frac{i^4}{4!}+\frac{i^5}{5!}+\cdots \\ &= 1+\frac{i}{1!}+\frac{-1}{2!}+\frac{-i}{3!}+\frac{1}{4!}+\frac{i}{5!}+\cdots \\ &=\left(1+\frac{-1}{2!}+\frac{1}{4!}+\cdots\right)+\left(\frac{i}{1!}+\frac{-i}{3!}+\frac{i}{5!}+\cdots\right) \\ &=\left(1-\frac{1}{2!}+\frac{1}{4!}+\cdots\right)+i\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}+\cdots\right) \\ &=\cos(1)+i\cdot\sin(1)\\ \end{align*}$$

OK, but isn't that just 1 step backwards in the proof of $e^x$'s MaClaurin expansion? — Meow, Jan 02 '13 at 21:12
@Alyosha There are so many different ways of getting at these formulae that I'm not sure the concepts of 'backwards' or 'forwards' even make sense here. That said, I don't know of any proof of that expansion that even involves sin/cos, so I'm not sure how it could be going backwards per se. — Steven Stadnicki, Jan 02 '13 at 21:27

Convergence of a spiral in $\mathbb{C}$

5 Answers5